Given $\mathfrak{q}$ in $L/K$ unramified, can we find $\alpha$ such that $L=K(\alpha)$ and $f'(\alpha)$ not divisible by $\mathfrak{q}$? Consider the following situation: Let $L/K$ be a finite extension of number fields and $\mathfrak{q}$ a prime of $L$ that is unramified. I am looking for a proof of the fact that one can always find some primitive element $\alpha$ of the field extension with the following property: If $f$ is the minimal polynomial of $\alpha$, then $f'(\alpha)\mathcal{O}_L$ is not divisible by $\mathfrak{q}$ - in other words, reducing everything mod $\mathfrak{q}$, $\overline{\alpha}$ is not a repeated root of $\overline{f}$.
I know that this is true "most of the time". Namely, if $\mathfrak{p}$ is the prime of $K$ below $\mathfrak{q}$, in the case where we can find some $\alpha$ with $L=K(\alpha)$ such that the conductor $\mathfrak{f}$ of $\mathcal{O}_K[\alpha]$ in $\mathcal{O}_L$ is coprime to $\mathfrak{p}\mathcal{O}_L$. However, this might not always be possible.
If it is indeed true, I would like to have a "global" proof of this fact, i.e. without localization.
(Background: I am trying to give a fairly simple proof that unramified primes $\mathfrak{q}$ do not divide the different $\mathcal{D}_{L/K}$. If one can always find an $\alpha$ as above, the argument is simple: We know $f'(\alpha)\in\mathcal{D}_{L/K}$ and hence $\mathcal{D}_{L/K}|f'(\alpha)\mathcal{O}_L$, which shows that $\mathcal{D}_{L/K}$ is not divisible by $\mathfrak{q}$.)
 A: Let $\mathfrak{p} =\mathfrak{q}\cap O_K$, $R=O_L/\mathfrak{p}O_L$, $k=O_K/\mathfrak{p}$, $\mathfrak{p}O_L=\mathfrak{q} I$.
$\mathfrak{q}$ unramified means that $I\not\subset \mathfrak{q}$ ie. $(\mathfrak{q},I)=(1)$ and $$R \cong O_L/\mathfrak{q}\times O_L/I\quad \text{   as } k \text{ algebra}\tag{1}$$

*

*Take $\beta\in I$ whose reduction generates $O_L/\mathfrak{q}$ as a field extension of $k$.


*Take $\gamma\in O_L$ that is a primitive element $L=\Bbb{Q}(\gamma)$.


*Let $$\alpha=\beta+n\gamma |R|$$ where $n$ is large enough such that $\forall \sigma,\sigma'\in Hom_\Bbb{Q}(L,\Bbb{C}), n |R||\sigma(\gamma)-\sigma'(\gamma)| > |\sigma(\beta)-\sigma'(\beta)|$.
This implies that $\Bbb{Q}(\alpha)$ has $[L:\Bbb{Q}]$ distinct complex embeddings ie. $L=\Bbb{Q}(\alpha)=K(\alpha)$.
Let $f(x)\in O_K[x]$ be $\alpha$'s monic minimal polynomial. Using $(1)$ we have
$$f(x) = \det(x-\alpha\in End_K(L))\equiv \det(x-\alpha\in End_k(R))\in k[x]$$
$$ = \det(x-\alpha\in End_k(O_L/\mathfrak{q}))\ \det(x-\alpha\in End_k(O_L/I))$$
$$= \det(x-\beta\in End_k(O_L/\mathfrak{q}))\ \det(x-0\in End_k(O_L/I)) = h(x) x^m\in k[x]$$
where $h(x)\in k[x]$ is separable and irreducible of degree $\dim_k(O_L/\mathfrak{q})$ and $m=\dim_k(O_L/I)$.
